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For let D E be the sine, and CE the versed sine of the arc DC, and since ab = 2 bc, by sim. triangles, AB= 2BC, but by the property of tangents to the circle, BC = BD, therefore, BD=DA, and by Simp. on Max. et Min. the rectangle DEX CE is a maxi.

Inference. The arc DC = 120°. For drawing the radius DO, since AB = 2BC, ABC is evidently half an equilateral triangle whose side is AB;, there. fore, the żB = 60°, the supplement of which is the angle DOC = arc DC = 120°. Otherwise by Messrs. Baines, Brooke, Cattrall, Cummins, Gawthorp, Hirst, Nesbit, Putsey, Rylando, and Webster,

Let radius = 1, and put the versed sine = 1; then, the sine =v(2x– xo), and <>(2 x — «) is to be a max.; make its fluxion = 0, and by reduction we have * = 1 = versed sine of 120°, the arc required. 10. Qu. (65) Answered by Mr. Vayheeg, the Proposer.

ANALYSIS. Suppose A the thing done, and that PAC is the circle


js is required; through P and O, the centre of the given circle, draw the right line CPEOF; then since AP, PD, PB is given, AP, PE, PF is given also. Por PD, PB = PE, PF, and because the point P and the circle EDBF are given in position, PE, PF is given; therefore, A P is equal to a given line, but one extremity P of it is given, consequently, the other extremity A must be in the circumference of a circle whose radius is AP. Consequently AP, PE, PP is equal to AP, PD, PB, the given solid.

Again, by Mr. Whitley. The point P and the circle EDBF being given, the rectangle PD, PB is given; and since the solid AP,

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PD, PB is given, AP is given; therefore, a circle de scribed about the centre P, with the given radius AP, will be that required.

Otherwise by Rylando. Let P be the given point, and QBD the

A А В. given circle; draw the


D tangent PQ, and produce it to S, so that QP. PS may be equal to the given solid; then with radius SP=PT=PA, and centre P, describe the circle STA, and the thing is done. For the given solid is QP”. PS = QPX PT = PA. PD. PB, by a known property of the circle.

In the same elegant manner it was answered also by Messrs. Brooke, Butterworth, Gawthorp, Hine, Maffetti Nesbit, Putsey, Tomlinson, and Winward. 11. Qu. (66) Answered by Messrs. Eyres and Rylando.

Let DR, CS, be the radii of P Р the given circles, and P a point in the required locus; from P draw the tangents PD, PC, and join PR, PS, RS; then, D it is evident that when PD+

R P+Ꭱ ” (PR) + PC + CS (PS), that is P R + PS’is constant; and because R and $ are constant the locus of P will be a circle.

Cor. 1. If the difference DP CP: be constant, the locus of P will be a right line perp. to RS.

Cor. 2. By Mr. Whitley. When R and S coincide, RS vanishes, so that, in this case, the locus of P is a given circle, concentric with the given circles; hence question the 18th (72) is only a particular case of the above.

This question was answered also by Messrs. Brooke, Butterworth, Gawthorp, Hine, Maffett, Nesbit, Putsey, Vayheeg, Whitley, and Winward.

31. asi 5T.

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AP"; therefore,

12. Qu. (67) Answered by the Proposer. Because AE. DC is, by the question, to be a max. and EDP, DAP, are right angles, AE =


A D2
* DC

D must be a max. or A D2

B x DC a

C max, which is known to be the case when AD=2DC. Hence we have the following

Construction. Divide AČ in D, such, that AD2 DC, join PD, and draw DE perp. to PD, intersecting BA produced in E, join EC, and the triangle EDČ will be a max. as required.

Otherwise by Mr. Rylando, Let A BC be the given triangle, P the given point, in the side A B, and draw the rest of the lines as in the figure above. Put AE=X, PA =m, and AC=p; then, AD= mx, the area of the triangle C AE = pr, and the area of the triangle DAE= ir mr; therefore, by the question 1px-4r./mx is a max. and its fluxion = 0, whence z is found = 4p, AD={p, and


2p3 the area of the triangle = To construct the

27 m problem; make AD=AC, join PD, and bisect it in m, draw mn perp. to PD, meeting PE in. n, with centre n, and radius Pn, describe a circle, which will cut PA produced in E, the point required.

Answers were also given by Messrs. Gawthorp, Hine, Maffett, Nesbit, Putsey, Whitley, and Winward. 13. Qu. (68) Answered by Mr. Nesbit, Farnley Academy.

Put the area, length and weight of one beam respectively A, L, and w, that of the other a, l, and w, and let S, s, G, g, be as in the question; then, by art. 225,

S X A XG Marrat's Mechanics,

LX W when the lengths are equal or L'=l, the weights as the areas, or W:w:: A: a, therefore,


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SX & Xg; but

x w


PAVA above equation, under these circumstances, becomes & XG=S,* g, that is, S:s::G:g. 2. E. D.

And in nearly the same manner it was answered by Messrs. Gawthorp, Harvey, Rylando, and Whitley. 14. Qu.(69) Answered by Mr. G. Harvey, jun. Plymouth. .

Analysis. Suppose the thing done, and that ABC is the tri

N angle required, because NC, DM the diff. of the seg, of the base, and NCF, the dift. of the angles at the base are given, HF, the diam. of the circumscribing cirele is given; and because LK=MC

A is given, ML drawn parallel to

EL-KB HF is given in position, and the centre of the inscribed circle is in this line, by the question ; but the centre of the in. scribed circle is known also to be in the line CF, it is therefore at S, their point of intersection.

Now', ' by prop. 35. Mod. Geom. Student. CS X SF =HF X SL and CS, SF, HF are all given; therefore LS is given; hence, if a circle be described from centre S with radius SL thus found, and CA, CB, be drawn from C, touching it, and cutting the circle ACB in A and B, and A, B, be joined, ABC will be the triangle required; as is evident from the analysis,

The same, by Messrs. Rylando and Whitley. ANALYSIS. Suppose the thing done, and that ABC is the triangle required; then, since the diff. of the angles at the base, and the diff. of the seg. of the base are given, the triangle CDF is given; draw CH perp. to CF, meeting FD produced in H, then HD, and the diam. HF are given, " But KL is given, and from thence EL; also by prop.:11 Mod. Geom. Student. EL = EK X EI; therefore, EI becomes known, and of course IK; but HDX DE=EK X IK; therefore, D E becomes : known, and the method of construction is evident.

Solutions were given also by Messrs. Butterworthsan Eyres, Hine, Webster, and Winward.

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15. Qu.(70) Answered by Mr. Nesint, Farnley Academy.

In the annexed figure let HO denote the horizon, EQ...

B the equator, 69 the tropic of Cancer, N the north and S E the south pole, F the town of Farnley, and D the point in which the sun's upper


HK would appear to a person elevated to A, on June 21st, at midnight.

Now, from the co-latitude = 36° 13', substract the declination = 23° 28', and we obtain 12° 45', the depression of the sun's centre below the horizon at midnight, from which take 15' 47" + 33 = 48' 47", the semi-diameter and refraction, and add 9" for parallax, and we get 11° 56' 22' = LOC 69 = - ACD. Then, in the right-angled triangle A DC, right-angled at D; as the nat. co-sine of 11° 56' 22" CD=3985 :: rad. : AC = 4073.1141 nearly; from - which take 3985, and there remains AF = 88,1141 miles, the height to which a person must be elevated, to see the sun's upper limb at the given time.

Put CF=r=21040800 feet, the radius of the earth; AC ==21506042,448 feet, the distance fallen from; g=1644 feet, half the velocity or force at F; n = the velocity required, and t = the time of falling from A to F; then (see Dr. Hutton's Course, vol. ij.


335; Mar. rat's Mechanics, art. 665; or Vince's Flux. art. 81) v=V Agrx

- 5411 feet d


second. Again, with the radius AG=GC, describe the semicircle ABC; draw B F perpendicular to AC, and join BG; then, we have BG=10753021.224, and GĚ= 10287778.776, to find the angle BGF = 16° 54 56", and side BF = 3128751.6. We also find the ato ABC 33781691.4773; and, as 180°: 33781691.4773 3: 10° 54' 56" : 3174644.883 = the arc A B; then, per

AB+BF idem, ut supra, + =


8 8' 53", the whole time of falling to the surface.

CFS 173"

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